How hard is LeetCode 2736. Maximum Sum Queries?

LeetCode 2736. Maximum Sum Queries is rated Hard on LeetCode.

What is the time complexity of LeetCode 2736. Maximum Sum Queries?

The Python solution on this page runs in O((n + m) \times \log n + m \times \log m).

What is the space complexity of LeetCode 2736. Maximum Sum Queries?

The Python solution on this page uses O(n + m) auxiliary space.

Maximum Sum Queries — LeetCode 2736 Python Solution

HardStackBinary Indexed TreeSegment TreeArrayBinary SearchSortingMonotonic Stack

Problem: #2736
Pattern: Stack
Reading time: 7 min
Source: leetcode.com

The problem

You are given two 0-indexed integer arrays nums1 and nums2, each of length n, and a 1-indexed 2D array queries where queries[i] = [xi, yi]. For the ith query, find the maximum value of nums1[j] + nums2[j] among all indices j (0 <= j < n), where nums1[j] >= xi and nums2[j] >= yi, or -1 if there is no j satisfying the constraints.

Example

Input: nums1 = [4,3,1,2], nums2 = [2,4,9,5], queries = [[4,1],[1,3],[2,5]]
Output: [6,10,7]
Explanation: For the 1st query xi = 4 and yi = 1, we can select index j = 0 since nums1[j] >= 4 and nums2[j] >= 1. The sum nums1[j] + nums2[j] is 6, and we can show that 6 is the maximum we can obtain.

Python solution

Python

class BinaryIndexedTree:
    __slots__ = ["n", "c"]

    def __init__(self, n: int):
        self.n = n
        self.c = [-1] * (n + 1)

    def update(self, x: int, v: int):
        while x <= self.n:
            self.c[x] = max(self.c[x], v)
            x += x & -x

    def query(self, x: int) -> int:
        mx = -1
        while x:
            mx = max(mx, self.c[x])
            x -= x & -x
        return mx


class Solution:
    def maximumSumQueries(
        self, nums1: List[int], nums2: List[int], queries: List[List[int]]
    ) -> List[int]:
        nums = sorted(zip(nums1, nums2), key=lambda x: -x[0])
        nums2.sort()
        n, m = len(nums1), len(queries)
        ans = [-1] * m
        j = 0
        tree = BinaryIndexedTree(n)
        for i in sorted(range(m), key=lambda i: -queries[i][0]):
            x, y = queries[i]
            while j < n and nums[j][0] >= x:
                k = n - bisect_left(nums2, nums[j][1])
                tree.update(k, nums[j][0] + nums[j][1])
                j += 1
            k = n - bisect_left(nums2, y)
            ans[i] = tree.query(k)
        return ans

Complexity

Measure	Complexity
Time	O((n + m) \times \log n + m \times \log m)
Space	O(n + m) auxiliary

Pattern: Stack

When the most recent unresolved thing is the one that matters, use a stack. LeetCode 2736. Maximum Sum Queries is filed here because LeetCode tags it Stack, which is the vocabulary this hub collects.

The stack guide has the Python template for the pattern and the 194 LeetCode problems that use it.

Frequently asked questions

How hard is LeetCode 2736. Maximum Sum Queries?: LeetCode 2736. Maximum Sum Queries is rated Hard on LeetCode.
What is the time complexity of LeetCode 2736. Maximum Sum Queries?: The Python solution on this page runs in O((n + m) \times \log n + m \times \log m).
What is the space complexity of LeetCode 2736. Maximum Sum Queries?: The Python solution on this page uses O(n + m) auxiliary space.
What topics does LeetCode 2736. Maximum Sum Queries cover?: LeetCode 2736. Maximum Sum Queries is tagged Stack, Binary Indexed Tree, Segment Tree, Array, Binary Search, Sorting and Monotonic Stack on LeetCode.